A projectile is shot from the ground at an angle of #pi/6 # and a speed of #25 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Question

A projectile is shot from the ground at an angle of #pi/6   # and a speed of #25   m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

William Audy · Accepted Answer

#~~31.2m#

The velocity of projection of the projectile #u=25ms^-1#
The angle of projection of the projectile #alpha=pi/6#
The horizontal component of velocity of projection
#u_h=ucosalpha=25xxcos(pi/6)=12.5sqrt3ms^-1#
TheVertical component of velocity of projection
#u_v=ucosalpha=25xxsin(pi/6)=12.5ms^-1#

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Assuming the ideal situation where gravitational pull is the only force acting on the body and no air resistance exists , we can easily proceed for various calculation using equation of motion under gravity.

CALCULATION
Let the projectile reaches its maximum height H m after t s of its start.
The final vertical component of its velocity at maximum height will be zero
So we can write
#0 = usinalpha-gxxt =>t = (usinalpha)/g=12.5/10=1.25s# taking #g=10ms^-2#

Again
#0^2=(usinalpha)^2-2xxgxxH #
#=>H = (usinalpha)^2/g=(12.5)^2/10=15.625m#

During # t=1.25s# of its ascent its horizontal component will remain unaltered and that is why the horizontal displacement
#R = ucosalpha xxt=12.5sqrt3xx1.25m=27m#

Hence the projectil's distance ( D ) from starting point to the maximum point of its ascent is given by
#D=sqrt(R^2+H^2)=sqrt(27^2+(15.625)^2)~~31.2m#

Please comment

Olivia Askew · Answer

undefined To find the distance from the starting point when the projectile reaches its maximum height, you can use the equation: $ d = v_0^2 \sin(2\theta) / g $, where $ v_0 $ is the initial velocity, $ \theta $ is the launch angle, and $ g $ is the acceleration due to gravity (approximately $ 9.8 \, \text{m/s}^2 $). Plugging in the given values: $ v_0 = 25 \, \text{m/s} $ and $ \theta = \pi/6 $, you get: $ d = (25^2) \sin(2(\pi/6)) / 9.8 $. Calculate $ \sin(2(\pi/6)) $ to find the answer.